In Freundlich adsorption isotherm, the value of \(\frac{1}{n}\) is:

between 0 and 1 in all cases

The correct answer is option 3. between 0 and 1 in all cases.

The value of \(\frac{1}{n}\) in the Freundlich adsorption isotherm is between 0 and 1 in all cases

The Freundlich isotherm is an empirical equation that describes the relationship between the amount of gas adsorbed by a solid (adsorbent) and the pressure of the gas (adsorbate) at a constant temperature. It's represented by the equation:

\[\frac{x}{m} = kP^{\frac{1}{n}}\]

where:

\(\frac{x}{m}\) is the amount of gas adsorbed per unit mass of adsorbent.

\(k\) and \(n\) are constants specific to the adsorbent-adsorbate system at a particular temperature

\(P\) is the pressure of the gas

The value of \(\frac{1}{n}\) determines the favorability of adsorption:

When \(0 < \frac{1}{n} < 1\), the adsorption increases with increasing pressure but at a decreasing rate. This is generally observed for physisorption, where there's a weak attraction between the gas molecules and the adsorbent surface.

If \(\frac{1}{n} = 0\), the adsorption becomes independent of pressure, which is a rare case.

If \(\frac{1}{n} = 1\), the adsorption is directly proportional to pressure. This is also uncommon.

Therefore, the value of \(\frac{1}{n}\) lies between \(0\) and \(1\) for most Freundlich isotherm applications, regardless of whether it's chemisorption (strong chemical bonding) or physisorption (weak physical attraction).

Chemisorption vs. Physisorption and Freundlich Isotherm:

The Freundlich isotherm doesn't inherently distinguish between chemisorption and physisorption. It describes the overall adsorption behavior. However, the value of \(\frac{1}{n}\) can sometimes provide clues:

Chemisorption typically involves a high energy adsorption process, leading to a stronger affinity for the surface. This might be reflected in a value of \(\frac{1}{n}\) closer to 1 (though not always).

Physisorption generally has a lower value of \(\frac{1}{n}\) due to the weaker interaction between the gas molecules and the surface.

It's important to remember that the Freundlich isotherm is an empirical model, and the interpretation of \(\frac{1}{n}\) should be done cautiously.